Here is an expanded version of my previous comments. There are a lot of things to check here and I haven't.
In my opinion the right axioms for Euclidean/Hyperbolic geometry are the Tarski axioms. Tarski works in a system where the domain is $\mathbb{R}^2$ and you have two relations, a ternary betweenness relation and an equidistance relation $E(x,y;x',y')$ which says that the line segments $\overline{xy}$ and $\overline{x',y'}$ are congruent. In this system all the axioms are phrased in terms of points, not in terms of both points and lines like in Euclid. Lines emerge as definable sets.
The Tarski axioms for Euclidean geometry are on Wikipedia, and you just change the parallel postulate to get axioms for Hyperbolic geometry. The usual proof of bi-interpretability goes by showing that both theories are bi-interpretable with the theory of $(\mathbb{R},+,×,<)$, i.e. the theory of real closed fields. I'm not sure where the hyperbolic case was written down, maybe by Szmielew. I don't know of a synthetic version of the proof.
But it should be possible to get a synthetic proof. There are well-known Euclidean models of the hyperbolic plane, like the Klein model, so it should be enough to make a Hyperbolic model of the Euclidean plane.
Edit: I think that my initial attempt at the Hyperbolic model of the Euclidean plane fails, because equidistance isn't definable. Following a suggestion of @ColinMcLarty, I will describe a different model that I think works. This model is from Greenberg's book "Euclidean and Non-euclidean geometries".
Let $\mathbb{H}$ be the Hyperbolic plane. In this model the points are just the points in $\mathbb{H}$ and the lines all all lines in $\mathbb{H}$ through the origin together with the curves in $\mathbb{H}$ that are equidistant from a line through the origin.
To get a model of Euclidean geometry we need a betweenness and equidistance relationship, and these relationships need to be definable in $\mathbb{H}$. The betweenness relationship is easy, you just need to observe that the lines form a uniformly definable family of sets. Equidistance is more complicated.
Greenberg describes a map $\rho : \mathbb{H} \to \mathbb{R}^2$ and says that the equidistance relation on $\mathbb{H}$ is the pull-back of the equidistance relation on $\mathbb{R}^2$ by $\rho$. We can put polar coordinates on $\mathbb{H}$ in the same way as on the Euclidean plane. We fix a ray $\ell$ through the origin and let the polar coordinates of $p \in \mathbb{H}$ be $(r,\theta)$ where $r$ is the Hyperbolic distance from the origin to $p$ and $\theta$ is the angle that $p$ makes with $\ell$. If $p \in \mathbb{H}$ is the point with polar coordinates $(r,\theta)$ then $\rho(p)$ is the point
$$
\rho(p) = (\sinh r \sin \theta, \sinh r \cos \theta) = \sinh r (\cos \theta, \sin \theta).
$$
So $\rho(p)$ is the point in $\mathbb{R}^2$ whose (euclidean) polar coordinates are $(\sinh r, \theta)$.
Now let's suppose that $\mathbb{H}$ is the Poincare disc model. I think it is enough to show that $\rho$ is semialgebraic, i.e. definable in $(\mathbb{R},+,\times,<)$. The Hyperbolic distance $r$ of $p$ from the origin is not a semialgebraic function of $p$, but it is the arcsinh of a semialgebraic functions, see Wikipedia. (On Wikipedia there is a factor of $2$ in from the arcsinh, but that is just a scaling factor, so we can drop it). I think it should be pretty clear that $\rho$ is semialgebraic.
Once we know that $\rho$ is semialgebraic we know that the equidistance relation on our Hyperbolic model of Euclidean geometry is semialgebraic. This should imply that it is definable in Hyperbolic geometry, but one would need extra arguments to see it directly.
Best Answer
A crucial difference between non-Euclidean geometries and "non-inductive" models of PA- is that any model of PA- contains a canonical copy of the true natural numbers, and in this copy of $N$, the induction schema is true. In other words, PA is part of the complete theory of a very canonical model of PA-, and as such it seems much more natural (so to speak) to study fragments of PA rather than extensions of PA- which contradict induction.
To make this a little more precise (and sketch a proof), the axioms of PA- say that any model $M$ has a unique member $0_M$ which is not the successor of anything, and that the successor function $S_M: M \to M$ is injective; so by letting $k_M$ (for any $k \in N$) be the $k$-th successor of $0_M$, the set $\{k_M : k \in N\}$ forms a submodel of $M$ which is isomorphic to the usual natural numbers, $N$. Any extra elements of $M$ not lying in this submodel lie in various "Z-chains," that is, infinite orbits of the model $M$'s successor function $S_M$.
(In the language of categories: the "usual natural numbers" are an initial object in the category of all models of PA-, where morphisms are injective homomorphisms in the sense of model theory.)
So, while PA seems natural, I'm not sure why there would be any more motivation to study PA- plus "non-induction" than there is to study any of the other countless consistent theories you could cook up, unless you find that one of these non-inductive extensions of PA- has a particularly nice class of models.